Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include<cstdio>
#include<algorithm>
#define N 200000
using namespace std;

int v[N+1];
int poz[N+1];
int dp[N+1];
int verp[N+1];

bool meow(int a,int b){
    return (poz[a]<poz[b]);
}

int main(){
    int n,i,j;

    scanf ("%d",&n);

    for(i=1;i<=n;i++){
        scanf ("%d",&v[i]);
        poz[v[i]]=i;
    }

    dp[n]=0;
    for(i=n-1;i>0;i--){
        if (dp[i+1]==0){
            if (poz[i]<poz[i+1]) dp[i]=0;
            else {
                dp[i]=1;
                verp[i]=i+1;
            }
        }
        else {
            v[0]=i;
            v[1]=i+1;
            v[2]=verp[i+1];

            sort(v,v+3,meow);
            for(j=0;v[j]!=i;j++);

            if (v[(j+1)%3]==i+1){
                dp[i]=dp[i+1];
                verp[i]=verp[i+1];
            }
            else {
                dp[i]=dp[i+1]+1;
                verp[i]=i+1;
            }
        }
    }

    printf ("%d",dp[1]);
    return 0;
}